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我对编程很陌生，我写了一个应用程序来处理特定形状的区域。

如果需要，我可以AreaCalculations在另一个班级提供，但效果很好。

我的问题是检查用户何时键入字符而不是双精度字符。正如您从我的代码中看到的那样，我通过使用 while 循环和(!reader.NextDouble()).

这可行，但我必须重复这个问题。我可以在整个程序中这样做，但有没有更简单/更整洁的方法？

谢谢，

克雷格

到目前为止我的代码：

package areaprog;

import java.util.Scanner;

public class Mainprog {


public static void main (String [] args){

    //Area Menu Selection
    System.out.println("What shape do you need to know the area of?\n" +
    "1: Square?\n" +
    "2: Rectangle?\n" +
    "3: Triangle?\n" +
    "4: Circle? \n" +
    "5: Exit\n"     
    );

    //User input for menu

    Scanner reader = new Scanner(System.in);
    System.out.println("Number: ");

    //Menu syntax checking
    while (!reader.hasNextDouble())
    {
        System.out.println("Thats not a number you tool.\n");
        System.out.println("Now pick again\n" +
                "1: Square?\n" +
                "2: Rectangle?\n" +
                "3: Triangle?\n" +
                "4: Circle? \n" +
                "5: Exit\n"     
                );

        reader.next(); //ask for next token?        
    }               
        double input = reader.nextDouble();
        reader.nextLine();



    //Depending on user selection, depends on what method is called using switch.
Scanner scan = new Scanner(System.in);

    //Square selection
    if (input == 1){
        System.out.println("What is a length of 1 side of the Square?\n");
            double s1 = scan.nextDouble();
            double SqAns = AreaCalculator.getSquareArea(s1);
            System.out.println("The area of you square is: " + SqAns);

                   }

    //Rectangle selection    
        if (input == 2){
        System.out.println("What is the width of your rectangle?.\n");
            double r1 = scan.nextDouble();
        System.out.println("What is the height of your rectangle?\n");
            double r2 = scan.nextDouble();
            double RecAns = AreaCalculator.getRectArea(r1, r2);
        System.out.println("The area of your rectangle is: " + RecAns);    
        }
    //Triangle selection
    if (input == 3){
        System.out.println("What is the base length of the triangle?.");
            double t1 = scan.nextDouble();
        System.out.println("What is the height of your triangle?");
            double t2 = scan.nextDouble();
            double TriAns = AreaCalculator.getTriArea(t1, t2);
        System.out.println("The area of your triangle is " + TriAns);
    }
    //Circle selection
    if (input == 4){
        System.out.println("What is the radius of your circle?.");
            double c1 = scan.nextDouble();
            double CircAns = AreaCalculator.getCircleArea(c1);
        System.out.println("The area of your circle is " + CircAns);    

    }
    //Exit application
    if (input == 5){
        System.out.println("Goodbye.");

    }


}

}

好的，所以我添加了一个异常来捕获错误。所以它现在在处理不使用整数的人的方式上更干净了。

数字：1 正方形的 1 边的长度是多少？

a 你为什么要变得聪明？使用整数。

但是然后程序就结束了....我如何让他们回到主菜单，甚至让他们重新输入最后的努力？

谢谢，

克雷格

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1 回答 1

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就菜单驱动程序而言，它很好。但正如你所说： “这有效，但我必须重复这个问题。”

所以阅读java中的异常处理教程

因此，您的代码将如下所示：

    try
   {
    System.out.println("Thats not a number you tool.\n");
    System.out.println("Now pick again\n" +
            "1: Square?\n" +
            "2: Rectangle?\n" +
            "3: Triangle?\n" +
            "4: Circle? \n" +
            "5: Exit\n"
            );


input = reader.nextDouble();
 }
   catch(InputMismatchException err)
    {
  System.out.print(err.getMessage());

     }

不要忘记导入import java.util.InputMismatchException或import java.util.*

并且尝试在try块外声明变量，可能在外面是不可见的。

于 2013-01-15T09:17:30.150 回答